Question

a bullet fired from a rifle attains a maximum height H of 25 m and a horizontal range of 200 m. Find the angle of projection.

Please solve this on a paper and post image ​

Answer 1

Given:

$\sf :\to Maximum \: height,h = \dfrac{ {u}^{2} { \sin}^{2} \theta }{2g} = 25m \\ \\ \\ \sf :\to Maximum \: range,R = \dfrac{ {u}^{2} \sin2\theta }{g} = 200m$

Find:

$\sf :\to angle \: of \: projection$

Solution:

$\sf :\implies Maximum \: height,h = \dfrac{ {u}^{2} { \sin}^{2} \theta }{2g} = 25m ....(i) \\ \\ \\ \sf :\implies Maximum \: range,R = \dfrac{ {u}^{2} \sin2\theta }{g} = 200m....(ii) \\ \\ \\ \\ \sf :\to dividing \: eq(i) \: by \: (ii) \: we \: get, \\ \\ \\ \sf : \implies \dfrac{ {u}^{2} { \sin}^{2} \theta }{2g} \times \dfrac{g}{ {u}^{2} \sin2\theta } = \frac{25}{200} \\ \\ \\ \sf : \implies \frac{ { \sin}^{2} \theta }{2 \sin 2 \theta} = \frac{1}{8} \\ \\ \\ \sf : \implies 4 { \sin}^{2} \theta = \sin 2 \theta \\ \\ \\ \sf : \implies 4\sin\theta \times \sin\theta = 2\sin\theta \times \cos \theta \\ \\ \\ \sf : \implies \frac{ \sin \theta}{\cos \theta} = \frac{1}{2} \\ \\ \\ \sf : \implies \tan \theta = \frac{1}{2} \\ \\ \\ \sf : \implies \theta = { \tan}^{ - 1} ( \frac{1}{2} ) \\ \\ \\ \sf : \implies 26 \degree-34'$

$\underline{\underline{\sf\therefore required \: angle \: of \: projection \: is \: 26 \degree-34'}}$